Question

As we all know, a char is 8 bit so it can store values from 0 to 255,

if I initialize a char with a greater value like 369, I get the same binary pattern than 113
i.e:01110001 under MSVC12

Is it a standard, or is it undefined behavior ?

Answer 1

Notice that 369 - 256 == 113.

4.7 Integral conversions, paragraphs 2 & 3:

If the destination type is unsigned, the resulting value is the least unsigned integer congruent to the source integer (modulo 2^n where n is the number of bits used to represent the unsigned type).

If the destination type is signed, the value is unchanged if it can be represented in the destination type (and bit-field width); otherwise, the value is implementation-defined.

But whether char means the same as unsigned char or signed char is also implementation-defined.

So this behavior does depend on the compiler, although most will do it this way.